Communication Systems (ECE 301): Analysis of Sample Exam Questions

This document presents a detailed analysis of sample exam questions covering various aspects of communication systems. The questions address topics such as the advantages and disadvantages of fiber optics and copper wires, the characteristics of digital signals, and the challenges of mobile network connectivity. Further, the document explores the average quantization error in A-D converters, minimum sample rates for audio codecs, and frequency-shift keying. It delves into the workings of synchronous TDM and asynchronous transmission, along with data compression techniques, including run length encoding and index compression. The document also touches on asymmetric cryptography, CSMA/CA, and the concept of ISMA. Finally, it includes calculations for maximum data carrying capacity and discusses the application of these concepts in real-world scenarios.

Sample exams questions 1
SAMPLE EXAMS QUESTIONS
By Name
Course
Instructor
Institution
Location
Date

Sample exams questions 2
Question 1 a.
a).
Advantages of fiber-optics over copper in communication media.
1. Greater bandwidth
Fiber –optics have relatively more bandwidths as compared to copper. This is because copper
wires were originally designed for a voice transmission hence lower bandwidth. Fiber optic
wires offer higher bandwidth for conveying higher data than copper wires of the equal sizes.
2. Reduced Security Risk:
The progression of the fiber optic communiqué sale is essentially compelled by aggregate
alertness in the data security anxieties and employment of the substitute raw material.
3. Longer Distance:
For the fiber optic broadcast, optical wires are able of offering less power loss that permits
signals to be conveyed to a more distance than for the copper wires.
4. Noise:
Practically transmission wires utilization, it’s unavoidable to achieve environments parameters
such as heating, ventilating, power substations among other industrialized sources of noise.
Nevertheless, fiber has a smaller rate of bit error of 10×10-13, as an effect of fiber of very
resilient to electromagnetic disturbances of transmission of Fiber optic is hence essentially
noise free as compared to copper.
Advantages of copper over fiber-optics.

Sample exams questions 3
5. The Cost is higher than in Copper wires:
Notwithstanding the truth is that optic fiber connection costs are falling by around 60% yearly,
connecting optic fiber wiring is comparatively higher than the cables for copper. This is true
since copper cable connection doesn’t require additional care like fiber cables.
b)
Digital signals are a sequences of pulsates with two states - on (illustrated by the logic ‘1’) or off
(illustrated by the logic ‘0’). Digital signals convey more data per second as compared to
analogue signals and hence they keep their quality of signal better over long distances of
transmission of signal. Therefore, digital signals are better in transmission of signal as the noise
is relatively lower as compared to the analogue signal as seen in the diagram below.
Fig 1: Showing the noise signal in the digital signal.
c)

Sample exams questions 4
1. It is really annoying as the mobile may indicate "bars" of network and indicate a good
connection of network, yet it is still incapable to do some very significant tasks.
2. Disappointingly, the Droid X hotspot and, to a slighter level, the 4G hotspot Epic were
not that easy to use.
d)
The channel connection in GSM between two mobile subscribers simplex is a full duplex since
for this system, these parties are capable to connect with each other concurrently. A prototype of
a full-duplex device is a cell phone; the individual at the two ends of a call are able to speak and
can be heard by the other individual concurrently (Max, 2014).
Question 2a
A)
The average quantization error in an A-D converter is ∆V/4 since during the sampling stage, a
full wave which has 2 cycles is used in the conversion. These two cycles will have 4 voltage
point hence the error can easily be obtained through ∆V/4.
b)

Sample exams questions 5
Minimum sample rates is 480 Hz for the A Codec for a mobile phone is required to convert
speech in the range 400Hz- 2kHz into a digital data stream.
C)
Frequency-shift keying permits digital data to be conveyed by variations or swings in the
frequency of a carrier signal, most normally it is an analog sine wave which is also a carrier
wave. There are always 2 binary states in a signal, logic one (1) and logic zero (0), each of these
are represented by an analog wave form (Johns, 2013).
d)
Synchronous TDM operates through the muliplexor hence resulting to accurately the equal
amount of time to every device connected to it. This duration of time portion is allotted even if a
device doesn’t have something to transmit. This is uneconomical since there will be several time
durations if the allocated time apertures are not being employed. While Time-division

Sample exams questions 6
multiplexing is a technique of placing several data streams in a sole signal through unraveling
the signal into several segments, every one has a very short duration of time.
Advantages of Synchronous
This reduces overhead bits
Advantages of adaptive time division multiplexing
TDM circuitry is not very complex
Question 3 a
a)
It helps in solving the problem of little frequency range (Bandwidth). This is the punishment for
acquaint with transitions of frequency. For a LAN of 10 Mbps, the spectrum signal which is
between the 5 and 20 MHz. The data word 111001 would look as the below;
b)
g/Symbol Frequency of Occurrence

Sample exams questions 7
[per 120 symbols]
# 34
@ 28
& 22
% 16
£ 12
$ 8
34 28 '@' 22 '&' 16 '%' 12 '£' 8 '$'
/ \
2 'b' 7 'c'

Sample exams questions 8
Now, repeat this process until only one tree is left:
34
/ \
9 12 'a' 13 'd' 14 'e' 8 'f'
/ \
2 'b' 7 'c'
21 27
/ \ / \
9 12 'a' 13 'd' 14 'e' 85 'f'
/ \
2 'b' 7 'c'
48
/ \

Sample exams questions 9
21 27
/ \ / \
9 12 'a' 13 'd' 14 'e' 85 'f'
/ \
2 'b' 7 'c'
133
/ \
48 85 'f'
/ \
21 27
/ \ / \
9 12 'a' 13 'd' 14 'e'
/ \
2 'b' 7 'c'
C)
c) Actions performed by Sender and Receiver
In cryptography of asymmetric, there are receiver and sender and each of them contain a key set
which is remote and the other one is public. The actions performed by both of them include;

Sample exams questions 10
Private-public keys
The sender will encrypt the message using its remote key when he wants to send a message then
he uses the communal key from the encrypts and receiver o'er.
Message Encrypted
The message which was encrypted is then conveyed to the receiver. The receiver uses its private
key to decrypt the encryption of outside on the message and then senders public key to decrypt
the message which was encrypted with the sender’s private key.
Message Decrypted
The original message is received by the receiver and this is how asymmetric cryptography in
different ways.
Digital Signature
The receiver gets the original document and is capable of proving that the sender since
unlocking of the message is done by public keys from the sender and this proves that the identity
of the person who sends the message.
d)
During the early 1970s there was key public progresses. The publication of the draft Data
Encryption Standard which was undertaken in the U.S. Federal Register on 17 March 1975. The

Sample exams questions 11
publication and the effort of coming up with a secure electronic communication demanded a lot
of finances which was not available at that time
Question 4a
a)
P-Persistent Carrier-sense multiple access is a media access control (MAC) protocol where a
node confirms the nonappearance of other traffic afore conveying on a shared transmission
medium like an electrical bus
b)
i. Gauge min and max values
Example
Taking the example of age as a variable, the aim should be to ensure that my minimum value is
sensible. Additionally, maxing should not be done somewhere in the region of 400’s.
ii. Look for missings.
Example
Columns are sorted in both ascending and descending order to identify the missing values in the
columns. Additionally, one can filter their dataset to only look at records that have a missing
value.
iii. Check the values of categorical variables
Example:

Sample exams questions 12
Correcting the categories of a spreadsheet from the upper part down to the lower part thus
making it easier to detect errors early enough
iv. Look at the ‘incidence rate’ of binary variables
Example,
Having 1’s and nulls instead of 1’s and 0’s which is easier to identify since the rate of the binary
variable will equate to 1.
c)
Question 5a
a) Calculating the maximum data carrying capacity of the cable
C= B log (1 + SNR)
C=?
B= 150 MHz
SNR= 1000
C= B log (1 + SNR)

Sample exams questions 13
150MHz log (1+ 1000)
150MHz x 9.967
= 1495. 05 Mbps
b)
In Carrier Sense Multiple Access/Collision Avoidance (CSMA/CA), everyone listens to the
inbound channel to see if they can transmit. The aim is to avoid collision therefore there is no
transmission of new packets if the inbound channel is still busy. This therefore demands that all
terminals of mobile receive signals from one another on the inbound frequency. However, some
terminals fail to sense a transmission by another terminal in mobile radio nets that have path
losses and fading channels.
Solution
The problem of the hidden terminal is avoided by using Inhibit Sense Multiple Access (ISMA).
In ISMA, the base station inhibits all other terminals from transmitting by transmitting a busy
signal on the outbound channel. This happens as soon as an inbound packet is received.
c)
Implementation of run length encoding
Actual software

Sample exams questions 14
This involves releasing a run-length encoding under the LGPL. Furthermore, fix bugs and
enhancements are added and they are posted as a newer version. The older version is also stored
for historical reference. Each of these versions is confined in a zipped archive that also has the
source files and some instructions on how to build an executable. Not even a single archive
contains executable programs. The type of information compressed with this method includes
simple graphic images, line drawings, and animations.
Implementation index compression
In each page that compression takes place, various steps are used by prefix compression;
 Selecting a value in each column which can be used to reduce storage space for the
values in each column
 Creation of row representing prefix values and stored in the compression information
structure which immediately follow the page header.
 The corresponding prefixes replace the values of prefixes which are repeated.
 But in case the row values are not matching, a partial match can still be indicated
The main type of information which is best compressed by this method is Read-only table and
partitions in Data warehouse environment.
Question 1 b
a)
Light is trapped in optical fibres by the use of TIR. The angle of incidence for the light entering
a fibre determines whether the light ,is guided by the fibre. Incident angle that is less than the
critical angle ensures that the light stays in the fiber. It is important to note that the index of

Sample exams questions 15
refraction for cladding must be lower than that of the core for the fibre to work. Cladding
protects the core from scratches and allows a narrow cone of light to be trapped in the fiber. The
range of angles in which light travels down the fibre is known as the cone of acceptance and its
apex angle is twice the cut-off angle.
The France-Corsica line (more than 300 km) used to be without repeater. However as the others
said, for long haul line, there is one EDFA each 80-100km (not that up to date, maybe new stuff
exist)
b)
Synchronous transmission is the process of transferring data which is characterized by a
continuous data streams in the form of signals which are complemented by regular timing
signals. These signals are generated by an external clocking mechanism whose aim is to
synchronize both the sender and the receiver with one another. Asynchronous transmission, on
the other hand, works in spurts and has to insert a start bit before each character of data and a
stop bit at its end to inform the receiver where it originates from and ends at. The term
asynchronous can therefore be said to mean a process where transmitted data is encoded with
start and stop bits to specify the beginning and end of each character.

Sample exams questions 16
Asynchronous is preferred since the additional bits provide the timing for the connection by
specifying when a complete character has been sent or received. Timing for each character
therefore, begins with the start bit and ends with the stop bit.
c)
1. A channel is one of multiple transmission paths within a single link between network points.
2. A channel is a separate path through which signals can flow.
Question 2 b
a)
Sampling
Sampling can be defined as the process of taking samples from the continuous time function.
Sampling is used to turn a continuous-time function into a discrete function that may take on
infinitely many values at different discrete indices. Sampling is normally done using Sample-
And-Hold circuit. A capacitor and a switch can however be used to perform simple experiments.
The sampling theorem is used when we intend to reconstruct a signal. The theorem states that a
sampling frequency two times the highest frequency that is expected is needed.
Quantization
Quantization is the process of mapping a continuous voltage signal to a discrete number of
voltage levels. The quantization noise is affected by the number of voltage levels. Because

Sample exams questions 17
digital computers are binary in nature, the number of quantization levels is usually a power of 2.
For example;
Where n is the number of quantization bits.
The signal may be amplified or attenuated before going into the analog to digital converter such
that the maximum and minimum levels of voltage give the best compromise between resolution
of the signal levels and minimization of clipping.
Encoding
Encoding is the process of digitally representing the quantized signal. It is performed by giving
each quantization level a unique label. For example, if four bits are used, the lowest level may be
(in binary) 0000, and the next highest level 0001.
b)

Sample exams questions 18
The average quantization error in an A-D converter is ∆V/4 since during the sampling stage; a
full wave which has 2 cycles is used in the conversion. These two cycles will have 4 voltage
points hence the error can easily be obtained through ∆V/4.
c)
Frequency-shift keying (FSK) is the process by which the frequency of a carrier signal is shifted
to allow the transmission of digital information. A signal has two binary states that are 1 and 0
each of which is represented by an analog wave form. A modem is used to convert the binary
data into an FSK signal that can then be transmitted via fiber optics, wireless media or telephone
lines.
Question 3b
a),
Symbol Frequency of Occurrence

Sample exams questions 19
(per 427 symbols)
* 61
^ 63
% 68
£ 104
$ 131
34 28 '@' 22 '&' 16 '%' 12 '£' 8 '$'
/ \
2 'b' 7 'c'
Now, repeat this process until only one tree is left:
34
/ \
9 12 'a' 13 'd' 14 'e' 8 'f'
/ \
2 'b' 7 'c'

Sample exams questions 20
21 27
/ \ / \
9 12 'a' 13 'd' 14 'e' 85 'f'
/ \
2 'b' 7 'c'
48
/ \
21 27
/ \ / \
9 12 'a' 13 'd' 14 'e' 85 'f'
/ \
2 'b' 7 'c'
133
/ \

Sample exams questions 21
48 85 'f'
/ \
21 27
/ \ / \
9 12 'a' 13 'd' 14 'e'
/ \
2 'b' 7 'c'
b)
Sender ‘A’ generates a key that he/she will to send communication to receiver ‘B’. But sender
‘A’ has to know receiver B’s public address that he/she will use to encrypt the message.
Therefore, receiver B transmits their public address to sender A via a reliable route. After getting
Receiver B’s public address, sender A uses it to encrypt the message and send it to the receiver.
The received message is then decrypted by receiver B using their private key.
Man in the middle attack is the type of attack where the attacker gets in between a
communication path. The attacker encrypts traffic between him/herself and the victim; he then
decrypts the message and reads the message. Here are two ways of countering this type of attack.
Firstly, never purchase or exchange sensitive information over a public Wi-Fi network.
Secondly, you need to have strong WAP and WEP encryption on all access points.
c),
It helps in solving the problem of little frequency range (Bandwidth). This is the drawback for
leading to frequent transitions. For a LAN of 10 Mbps, the signal spectrum are between the 5 and
20 MHz. The data word 111001 would look as the below;

Sample exams questions 22
Question 4b
a)
Frequency hopping is a situation in which the transmitter changes the frequency of the carrier signal
depending on a particular hoping pattern. During each hop, the signal is able to see different channels
and varying sets of interfering signals. The result is that communication failure at given frequencies due
to fading or interference is avoided. It can be applied in GSM telephony and wireless LANs. It is
characterized by the following two advantages. Firstly, it provides the same quality of communication
for all the users in the channel. Secondly, it enables users to get an agreeable radio environment.
b)
Voice over IP (VoIP) can be described as the process of transmitting voice multimedia content over
internet protocols. It makes use of codec to envelop audio into data packets, transfer the packets across
an IP network and unwrap the packets back to audio at the receiving end. Service providers use VoIP to
deliver services over private networks and their broadband. As a result, enterprises can be allowed to
operate a single data and voice network. It also enables fast failovers around redundant communication
and outages between several endpoints thus enabling service providers to overcome the challenge of
making a connection between a PC and a landline telephone.
C)
Hamming codes can be employed to locate and then correct a single bit error.
The 8-bit data word given is 00001111
To get the 12-bit hamming code we have to include 4 parity bits as shown below.

Sample exams questions 23
1 2 3 4 5 6 7 8 9 10 11 12
P1 P2 0 P4 0 0 0 P4 1 1 1 1
P1 Checks bits (1, 3, 5, 7, 9, 11)
Therefore, P1= XOR of bits (3, 5, 7, 9, 11)
¿ 0 ⨁ 0 ⨁ 0 ⨁ 1 ⨁1 ¿ 0
P2 Checks bits (2, 3, 6, 7, 10, 11)
Therefore, P2= XOR of bits (3, 6, 7, 10, 11)
¿ 0 ⨁ 0 ⨁ 0 ⨁ 1 ⨁1 ¿ 0
P4 Checks bits (4, 5, 6, 7, 12)
Therefore, P4 =XOR of bits (5, 6, 7, 12)
¿ 0 ⨁ 0 ⨁ 0 ⨁ 1 ¿ 1
P8 Checks bits (8, 9, 10, 11, 12)
Therefore, P8= XOR of bits (9, 10, 11, 12)
¿ 1 ⨁ 1 ⨁ 1⨁ 1 ¿ 0
The 12-bit hamming code is therefore,
000100001111

Sample exams questions 24
Question 5b
a)
Maximum data carrying capacity is calculated as shown below.
Where C is the carrying capacity, B is the bandwidth, S is the signal power and N is the noise power.
B=1000Hz
S/N=2
The signal to noise ratio is converted to dB as follows.
S/N= C/B
Therefore,
C
1000= 2
C= 2000
b)
GSM refers to Global System for Mobile communication. It employs both Frequency Division
Multiplexing (FDM) and Time Division Multiplexing (TDM). FDM divides a channel into two or more
ranges of frequency that do not overlap. TDM, on the other hand, divides and allocates particular
periods of time to each channel in an alternating manner. FDM enables GSM to use the 124 uplink
channels to transmit data from cell phones to the BTS towers. The other 124 downlink channels are
used to transmit data from the BTS antenna to the cell phones on the ground. Therefore, each signal
uses a portion of the bandwidth at all times. TDM divides the channels into 8 time slots thus enabling
each signal to have access of the whole bandwidth at a particular time. The communication link may
appear to be full-duplex since signals move between the cell phone and BTS towers almost
simultaneously. But the communication is in fact half-duplex and therefore communication can only go
one way at a particular time. This can be further supported by the fact that the channel is divided into
time slots and links such that at any particular time, signals will have access of one of the channels, that
is, either uplink or downlink implying that communication will be one way.

Sample exams questions 25
c)
The above figure shows a two-level and a four-level signaling. Multilevel signaling is used to compress
the required bandwidth to transmit data at a particular bit rate. In this type of signaling, the symbol rate
is normally less than the bit rate. For example, in the four-level scheme above, groups of data bits pairs
are mapped to one of the four symbols. However, only one symbol is transmitted for each pair of the
data bits. As a result the symbol rate is half of the bit rate. This scheme is however very susceptible to
noise interference.
QUESTION 1 C
a)
Advantages of fiber-optics over copper in communication media.
1. Greater bandwidth
Fiber –optics have relatively more bandwidths as compared to copper. This is because copper
wires were originally designed for a voice transmission hence lower bandwidth. Fiber optic
wires offer higher bandwidth for conveying higher data than copper wires of the equal sizes.
2. Reduced Security Risk:
The progression of the fiber optic communiqué sale is essentially compelled by aggregate
alertness in the data security anxieties and employment of the substitute raw material.

Sample exams questions 26
3. Longer Distance:
For the fiber optic broadcast, optical wires are able of offering less power loss that permits
signals to be conveyed to a more distance than for the copper wires.
4. Noise:
Practically transmission wires utilization, it’s unavoidable to achieve environments parameters
such as heating, ventilating, power substations among other industrialized sources of noise.
Nevertheless, fiber has a smaller rate of bit error of 10×10-13, as an effect of fiber of very
resilient to electromagnetic disturbances of transmission of Fiber optic is hence essentially
noise free as compared to copper.
Advantages of copper over fiber-optics.
5. The Cost is higher than in Copper wires:
Notwithstanding the truth is that optic fiber connection costs are falling by around 60% yearly,
connecting optic fiber wiring is comparatively higher than the cables for copper. This is true
since copper cable connection doesn’t require additional care like fiber cables.
c),
A handover is a procedure mobile communications as well as telecom in which a coupled data
session or cellular call is conveyed from one cell site to another minus detaching the session.
Question 2 c
a),

Sample exams questions 27
This is the error resultant of attempting to signify an endless analog signal with distinct and
stepped digital information. The problematic ascends if the value of the analog being sampled
goes down amid two digital “steps
b)
Highest frequency
Lowest frequency
The bandwidth
Therefore
Sampling rate
Sampling rate is therefore 100000 samples/sec
c),
Amplitude Shift Keying is a kind of Amplitude Modulation that signifies the binary data in away
of discrepancies in the amplitude of a signal.

Sample exams questions 28
d)
Frequency Division Multiplexing is a method of networking where the multiple signals are
joined for concurrent communication through a shared communiqué medium. FDM employs a
carrier signal at a distinct frequency for every data stream and then associates with several
modulated signals.
Code division multiplexing (CDM) is a networking technique where multiple data signals are
joint for concurrent communication over a common frequency band.
Advantage of FDM:
1. This method is highly accurate, since it uses fine mesh.
2. It is the only widely used method in time domain for solving general problems.
3. Simple to implement than MOM or FEM.

Sample exams questions 29
Disadvantage of FDM:
1. Meshing is inflexible
2. Uncertainty about precision of boundaries
3. Frequency domain finite difference formulation are available but never become popular
for general problems.
Advantages of CDMA techniques:
 Efficient practical utilization of fixed frequency spectrum.
 Flexible allocation of resources.
 Many users of CDMA use the same frequency, TDD or FDD may be used
 Multipath fading may be substantially reduced because of large signal bandwidth
 No absolute limit on the number of users, Easy addition of more users.
 Impossible for hackers to decipher the code sent
 Better signal quality
 No sense of handoff when changing cells
 The CDMA channel is nominally 1.23 MHz wide.
 CDMA networks use a scheme called soft handoff, which minimizes signal breakup as a
handset passes from one cell to another.
 CDMA is compatible with other cellular technologies; this allows for nationwide
roaming.
 The combination of digital and spread-spectrum modes supports several times as many
signals per unit bandwidth as analog modes.

Sample exams questions 30
Disadvantages to using CDMA:
 As the number of users increases, the overall quality of service decreases
 Self-jamming
 Near- Far- problem arises
Question 3 c
a)
Initial leaves {(%50) (*42) (&35) (E15) ($8) (^3)}
Merge {(%50) (*42) (&35) ({E $} 23) (^3)}
Merge {(%50) (*42) (&35) ({E $ ^} 26)}
Merge {(%50) ({* &} 77) ({E $ ^} 26)}
Merge {(%50) ({* & E $ ^} 103)}
Final Merge {({%* & E $ ^} 153)}
b)
Sender ‘A’ generates a key that he/she will to send communication to receiver ‘B’. But
sender ‘A’ has to know receiver B’s public address that he/she will use to encrypt the
message. Therefore, receiver B transmits their public address to sender A via a reliable
route. After getting Receiver B’s public address, sender A uses it to encrypt the message
and send it to the receiver. The received message is then decrypted by receiver B using
their private key.
Man in the middle attack is the type of attack where the attacker gets in between a
communication path. The attacker encrypts traffic between him/herself and the victim; he
then decrypts the message and reads the message. Here are two ways of countering this
type of attack. Firstly, never purchase or exchange sensitive information over a public
Wi-Fi network. Secondly, you need to have strong WAP and WEP encryption on all
access points.

Sample exams questions 31
C),
Question 4 c
a),
Bluetooth permits the connection of two devices to each other without using wire. The utmost
corporate use of Bluetooth technology is in hands-free devices like headphones employed with
mobile phones. Most people have successfully solve this problem by using cell phones
exclusively, this will highly help in reducing the problem.
b),
For the cellular idea, frequencies allotted to the service are always used again in a steady
configuration of regions, known as 'cells', every region is covered by one vile station. In

Sample exams questions 32
telephone nets these cells are generally hexagonal. But for the radio broadcasting, a comparable
idea has been established based on rhombic cells.
To confirm that the communal intervention between operators remains underneath a destructive
level, advancing cells employs several frequencies.
c),
The word received is 011100001000
The above word can be represented in a table as shown below
1 2 3 4 5 6 7 8 9 10 11 12
0 1 1 1 0 0 0 0 1 0 0 0
Where P1=0 , P2=1 , P4 =1, P8=0
Because of even parity C=C8 C4 C2 C1
From the parities above, C1=C8 C4 C2 C1=0110 present at the sender end
C2=C8 C4 C2 C1=0110 present at thereceiver end
But, C1 ⨁ C2 should be 0000 if there is no error
Thus C1 ⨁ C2=0110 ⨁ 0110=0000
Therefore, there is no error and the 8-bit data is
10001000

Sample exams questions 33
Question 5 c
a)
Maximum data carrying capacity is calculated as shown below.
Where C is the carrying capacity, B is the bandwidth, S is the signal power and N is the noise
power.
B=5MHz
S/N=99
The signal to noise ratio is converted to dB as follows.
Therefore,
S
N = C
B
99= C
5
C= 99× 5
C= 495
b),

Sample exams questions 34
Cycle-accurate model
Presentation examination can’t be done on an abstract representation of the SoC. Even though a
operation level model of an SoC is very helpful to aid early development of software of a cycle-
accurate model is needed for eloquent concert analysis.
2) Automatic RTL generation
RTL is constantly created as chunk of the SoC expansion process. Nevertheless, it is gradually
mutual for precise IP to be produced from conformation Meta data.
3) Verification IP
Concert verification needs seizing contacts which crosses the SoC components which are
interconnect. Verification IP (VIP) checks are required to lookout traffic at all the interconnect
ports.
4) Testbench generation
A testbench performance is required to drive useful simulation runs of the several SoC
conformations. The testbench should be altered for every SoC RTL conformation, hence a
generator is needed to hustle the procedure and make it practical to achieve presentation analysis.
5) In-depth analysis
when a cycle precise RTL model and testbench have been created, tests should be run alongside
the design and several transactions scrutinized through the VIP. Investigation solution is to have
a serious analysis structures. Graphical presentations are needed to envisage results and erudite
clarifying machineries are needed to recognize root causes.
C),

Sample exams questions 35
Implementation of run length encoding
Actual software
This will involve a release of run-length encoding under the LGPL. Adding fix bugs
and enhancements which will be posted as a newer version whereby the bigger the
variety number, the novel the version. The older version will be retained for historical
reasons
Every version is in its personal zipped file that comprises the source files and brief
instructions for building an executable. None of the archives contain executable program
and a copy of the archives is obtained from;
Version 0.4: Altered the API so that decode and encode routines admit undid file
stream in its place of file names.
Version 0.3: Doses bug for maximum dimension goes in the variation of the pack
bits’ algorithm. Obviously license the library in LGPL version 3
Version 0. 2: Comprises an optional of the pack bits’ algorithm
Version 0.1: Original release. It does not contain any pack bits’ style encoding
All the sources I had to provide written in stern ANSI C. I would presume it to develop
correctly on my machine with an ANSI C compiler and testing of code compiled with gcc
was done on Windows XP and Linus.
The type of information compressed with this method includes simple graphic images,
line drawings, and animations.

Sample exams questions 36
Implementation index compression
In each page that compression takes place, various steps are used by prefix compression;
 Selecting a value in each column which can be used to reduce storage space for the
values in each column
 Creation of row representing prefix values and stored in the compression information
structure which immediately follow the page header.
 The corresponding prefixes replace the values of prefixes which are repeated.
 But in case the row values are not matching, a partial match can still be indicated
The main type of information which is best compressed by this method is Read-only table
and partitions in Data warehouse environment.

Sample exams questions 37
Bibliography
Johns, D. (2013). Conversion of analog signals to digital signals . Chicago : Springer .
Max, k. (2014). Usage of bluetooth in trasnfering of information . Hull: CRC.